Mixing
Prove $a,b in G Rightarrow |ab| = |a^{-1} b^{-1}|$
$begingroup$
Suppose $|ab| = n$
Then we know $(ab)^n = e$ where $e$ is the identity element in the group $G$. Assuming it is not Abelian.
We want to show $|a^{-1}b^{-1}| = |ab| Rightarrow (a^{-1}b^{-1})^n = e$
I'm not quite sure where to begin.
What's a starting point I can start off with?
group-theory
asked Jan 29 at 16:52
user2965071user2965071
1887
$endgroup$
add a comment |
$begingroup$
Suppose $|ab| = n$
Then we know $(ab)^n = e$ where $e$ is the identity element in the group $G$. Assuming it is not Abelian.
We want to show $|a^{-1}b^{-1}| = |ab| Rightarrow (a^{-1}b^{-1})^n = e$
I'm not quite sure where to begin.
What's a starting point I can start off with?
group-theory
asked Jan 29 at 16:52
user2965071user2965071
1887
$endgroup$
$begingroup$
I think it should be $b^{-1}a^{-1}$ instead of $a^{-1}b^{-1}$. Or there might be an assumption about the group being Abelian or something.
$endgroup$
– stressed out
Jan 29 at 16:54
$begingroup$
@stressedout It is assumed that $G$ is not Abelian.
$endgroup$
– user2965071
Jan 29 at 16:57
$begingroup$
There is no reason to assume that $G$ is not abelian. The statement is true in any case.
$endgroup$
– Derek Holt
Jan 29 at 21:48
add a comment |
$begingroup$
Suppose $|ab| = n$
Then we know $(ab)^n = e$ where $e$ is the identity element in the group $G$. Assuming it is not Abelian.
We want to show $|a^{-1}b^{-1}| = |ab| Rightarrow (a^{-1}b^{-1})^n = e$
I'm not quite sure where to begin.
What's a starting point I can start off with?
group-theory
asked Jan 29 at 16:52
user2965071user2965071
1887
$endgroup$
Suppose $|ab| = n$
Then we know $(ab)^n = e$ where $e$ is the identity element in the group $G$. Assuming it is not Abelian.
We want to show $|a^{-1}b^{-1}| = |ab| Rightarrow (a^{-1}b^{-1})^n = e$
I'm not quite sure where to begin.
What's a starting point I can start off with?
group-theory
group-theory
asked Jan 29 at 16:52
user2965071user2965071
1887
asked Jan 29 at 16:52
user2965071user2965071
1887
edited Jan 29 at 16:56
user2965071
asked Jan 29 at 16:52
user2965071user2965071
1887
asked Jan 29 at 16:52
user2965071user2965071
1887
asked Jan 29 at 16:52
user2965071user2965071
1887
1887
$begingroup$
I think it should be $b^{-1}a^{-1}$ instead of $a^{-1}b^{-1}$. Or there might be an assumption about the group being Abelian or something.
$endgroup$
– stressed out
Jan 29 at 16:54
$begingroup$
@stressedout It is assumed that $G$ is not Abelian.
$endgroup$
– user2965071
Jan 29 at 16:57
$begingroup$
There is no reason to assume that $G$ is not abelian. The statement is true in any case.
$endgroup$
– Derek Holt
Jan 29 at 21:48
add a comment |
$begingroup$
I think it should be $b^{-1}a^{-1}$ instead of $a^{-1}b^{-1}$. Or there might be an assumption about the group being Abelian or something.
$endgroup$
– stressed out
Jan 29 at 16:54
$begingroup$
@stressedout It is assumed that $G$ is not Abelian.
$endgroup$
– user2965071
Jan 29 at 16:57
$begingroup$
There is no reason to assume that $G$ is not abelian. The statement is true in any case.
$endgroup$
– Derek Holt
Jan 29 at 21:48
$begingroup$
I think it should be $b^{-1}a^{-1}$ instead of $a^{-1}b^{-1}$. Or there might be an assumption about the group being Abelian or something.
$endgroup$
– stressed out
Jan 29 at 16:54
$begingroup$
I think it should be $b^{-1}a^{-1}$ instead of $a^{-1}b^{-1}$. Or there might be an assumption about the group being Abelian or something.
$endgroup$
– stressed out
Jan 29 at 16:54
$begingroup$
@stressedout It is assumed that $G$ is not Abelian.
$endgroup$
– user2965071
Jan 29 at 16:57
$begingroup$
@stressedout It is assumed that $G$ is not Abelian.
$endgroup$
– user2965071
Jan 29 at 16:57
$begingroup$
There is no reason to assume that $G$ is not abelian. The statement is true in any case.
$endgroup$
– Derek Holt
Jan 29 at 21:48
$begingroup$
There is no reason to assume that $G$ is not abelian. The statement is true in any case.
$endgroup$
– Derek Holt
Jan 29 at 21:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, observe that $;|x|=|x^{-1}|;$ for any $;xin G;$ , and thus
$$b^{-1}a^{-1}=(ab)^{-1}implies|b^{-1}a^{-1}|=|(ab)^{-1}|=|ab|$$
But also observe that
$$ba=a^{-1}(ab)aimplies |ab|=|ba|;text{ since conjugate elements have the same order}$$
Put the above together and prove your quest.
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
$endgroup$
3
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
add a comment |
$begingroup$
Hint:
$|ghg^{-1}|=|h|$
$|h^{-1}|=|h|$
Apply this to $h=ab$ and a suitable $g$.
answered Jan 29 at 17:00
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
$$ab . ab cdots ab = e$$
$$ b . ab cdots ab = a’$$ multiply from the left by a inverse
$$b.abcdots .a= a’b’$$multiply from the right by b inverse
$$b.abcdots ab.= a’b’a’ $$multiply from the right by a inverse keep going
answered Jan 29 at 17:02
AmeryrAmeryr
689311
$endgroup$
1
$begingroup$
Use$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$
$endgroup$
– Shaun
Jan 29 at 17:12
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, observe that $;|x|=|x^{-1}|;$ for any $;xin G;$ , and thus
$$b^{-1}a^{-1}=(ab)^{-1}implies|b^{-1}a^{-1}|=|(ab)^{-1}|=|ab|$$
But also observe that
$$ba=a^{-1}(ab)aimplies |ab|=|ba|;text{ since conjugate elements have the same order}$$
Put the above together and prove your quest.
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
$endgroup$
3
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
add a comment |
$begingroup$
First, observe that $;|x|=|x^{-1}|;$ for any $;xin G;$ , and thus
$$b^{-1}a^{-1}=(ab)^{-1}implies|b^{-1}a^{-1}|=|(ab)^{-1}|=|ab|$$
But also observe that
$$ba=a^{-1}(ab)aimplies |ab|=|ba|;text{ since conjugate elements have the same order}$$
Put the above together and prove your quest.
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
$endgroup$
3
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
add a comment |
$begingroup$
First, observe that $;|x|=|x^{-1}|;$ for any $;xin G;$ , and thus
$$b^{-1}a^{-1}=(ab)^{-1}implies|b^{-1}a^{-1}|=|(ab)^{-1}|=|ab|$$
But also observe that
$$ba=a^{-1}(ab)aimplies |ab|=|ba|;text{ since conjugate elements have the same order}$$
Put the above together and prove your quest.
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
$endgroup$
First, observe that $;|x|=|x^{-1}|;$ for any $;xin G;$ , and thus
$$b^{-1}a^{-1}=(ab)^{-1}implies|b^{-1}a^{-1}|=|(ab)^{-1}|=|ab|$$
But also observe that
$$ba=a^{-1}(ab)aimplies |ab|=|ba|;text{ since conjugate elements have the same order}$$
Put the above together and prove your quest.
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
answered Jan 29 at 16:59
DonAntonioDonAntonio
180k1494233
180k1494233
3
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
add a comment |
3
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
3
3
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
$begingroup$
The last line that you added after your edit was the part that was missing when I was trying to solve it. (+1)
$endgroup$
– stressed out
Jan 29 at 17:01
add a comment |
$begingroup$
Hint:
$|ghg^{-1}|=|h|$
$|h^{-1}|=|h|$
Apply this to $h=ab$ and a suitable $g$.
answered Jan 29 at 17:00
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Hint:
$|ghg^{-1}|=|h|$
$|h^{-1}|=|h|$
Apply this to $h=ab$ and a suitable $g$.
answered Jan 29 at 17:00
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Hint:
$|ghg^{-1}|=|h|$
$|h^{-1}|=|h|$
Apply this to $h=ab$ and a suitable $g$.
answered Jan 29 at 17:00
lhflhf
167k11172403
$endgroup$
Hint:
$|ghg^{-1}|=|h|$
$|h^{-1}|=|h|$
Apply this to $h=ab$ and a suitable $g$.
answered Jan 29 at 17:00
lhflhf
167k11172403
answered Jan 29 at 17:00
lhflhf
167k11172403
answered Jan 29 at 17:00
lhflhf
167k11172403
answered Jan 29 at 17:00
lhflhf
167k11172403
167k11172403
add a comment |
add a comment |
$begingroup$
$$ab . ab cdots ab = e$$
$$ b . ab cdots ab = a’$$ multiply from the left by a inverse
$$b.abcdots .a= a’b’$$multiply from the right by b inverse
$$b.abcdots ab.= a’b’a’ $$multiply from the right by a inverse keep going
answered Jan 29 at 17:02
AmeryrAmeryr
689311
$endgroup$
1
$begingroup$
Use$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$
$endgroup$
– Shaun
Jan 29 at 17:12
add a comment |
$begingroup$
$$ab . ab cdots ab = e$$
$$ b . ab cdots ab = a’$$ multiply from the left by a inverse
$$b.abcdots .a= a’b’$$multiply from the right by b inverse
$$b.abcdots ab.= a’b’a’ $$multiply from the right by a inverse keep going
answered Jan 29 at 17:02
AmeryrAmeryr
689311
$endgroup$
1
$begingroup$
Use$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$
$endgroup$
– Shaun
Jan 29 at 17:12
add a comment |
$begingroup$
$$ab . ab cdots ab = e$$
$$ b . ab cdots ab = a’$$ multiply from the left by a inverse
$$b.abcdots .a= a’b’$$multiply from the right by b inverse
$$b.abcdots ab.= a’b’a’ $$multiply from the right by a inverse keep going
answered Jan 29 at 17:02
AmeryrAmeryr
689311
$endgroup$
$$ab . ab cdots ab = e$$
$$ b . ab cdots ab = a’$$ multiply from the left by a inverse
$$b.abcdots .a= a’b’$$multiply from the right by b inverse
$$b.abcdots ab.= a’b’a’ $$multiply from the right by a inverse keep going
answered Jan 29 at 17:02
AmeryrAmeryr
689311
answered Jan 29 at 17:02
AmeryrAmeryr
689311
answered Jan 29 at 17:02
AmeryrAmeryr
689311
answered Jan 29 at 17:02
AmeryrAmeryr
689311
689311
1
$begingroup$
Use$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$
$endgroup$
– Shaun
Jan 29 at 17:12
add a comment |
1
$begingroup$
Use$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$
$endgroup$
– Shaun
Jan 29 at 17:12
1
1
$begingroup$
Use
$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$ for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$$endgroup$
– Shaun
Jan 29 at 17:12
$begingroup$
Use
$$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$ for $$underbrace{abcdotdotscdot ab}_{ntext{ times}.}$$$endgroup$
– Shaun
Jan 29 at 17:12
add a comment |
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$begingroup$
I think it should be $b^{-1}a^{-1}$ instead of $a^{-1}b^{-1}$. Or there might be an assumption about the group being Abelian or something.
$endgroup$
– stressed out
Jan 29 at 16:54
$begingroup$
@stressedout It is assumed that $G$ is not Abelian.
$endgroup$
– user2965071
Jan 29 at 16:57
$begingroup$
There is no reason to assume that $G$ is not abelian. The statement is true in any case.
$endgroup$
– Derek Holt
Jan 29 at 21:48